Some fractals have a whole fractal dimension, can their measure be calculated? For example if you start with a tetrahedron of a given size and recursively remove the central octahedron leaving 4 tetrahedrons of half the side length you will end up with a fractal of dimension 2. How does the area of this fractal relate to the size of the original tetrahedron? There is an image of this fractal here. What if you start with a cube, divide it into eight smaller cubes, remove the four cubes of a given parity and continue recursively as shown here? If you generate the Type 2 Quadratic curve (dimension 1.5) on one plane and a 0.5 dimensional Generalized Cantor Set on the perpendicular axis, produce a plane going through every point in the Cantor set and project the curve on each plane, what is the area of this?
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Hint: at every step, the surface area stays constant. – Wojowu Jan 08 '16 at 16:44
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I do not mean the surface area of the fractal, I mean the measure of the fractal itself. – k-l Jan 08 '16 at 17:21
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Also, for some similar fractals it does not stay constant. – k-l Jan 08 '16 at 20:33
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Also, if the four top cubes are always kept the final result is a plane with area l^2 (l is the side length of the original cube) but the total area of the cube is 6 l^2, which is much bigger. – k-l Jan 08 '16 at 21:43
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You say "Area" in your question, not "Measure"...also do you mean measure of the fractal or measure of the boundary? – Zach466920 Jan 09 '16 at 16:21
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I mean the measure of the fractal itself, I say area because the fractal dimension is 2. – k-l Jan 09 '16 at 17:11